Big review 4 finals (2 posts in 1) check my work

A 7.58 gram sample of benzoic acid, C6H5CO2H, is dissolved in 100 mL of benzene. The density of benzene is 0.879 g/mL. What is the molality of the benzoic acid?

Sol: Convert mL to grams benzene.

100 mL * (0.879g benzene/1 mL) = 87.9 g

Molality = # molsolute / # kg solvent

So I convert above answer to kg:
87.9 g * (1 kg/1000 g)=.0879 kg Benzene

M.W. Benzene: 122.12 g

?mol Benzene :
7.58g benzene*(1 mol benzene/122.12g)=0.0621 mol Benzene.

Molality =
.0621 mol/.0879 kg =.706 mol/kg benzene

2. What is the molarity of a Ba(OH)2 solution if 15.48 mL of it are required to react with 25 mL of .303 M HCl solution? Show the balanced reaction in your calculations.

Sol:
2HCl(aq)+ Ba(OH)2(aq)-->BaCl2 + 2H2O(L)

(there are some exceptions when dealing with the formula: m1v1 = m2v2; I know to apply when you deal with the same compound. Here we have two. I'll use formula anyways as I see no way to proceed.

(2*.3030M HCl)*.025L = .0155M2
.0152M = m2 ??







A 7.58 gram sample of benzoic acid, C6H5CO2H, is dissolved in 100 mL of benzene. The density of benzene is 0.879 g/mL. What is the molality of the benzoic acid?

Sol: Convert mL to grams benzene.

100 mL * (0.879g benzene/1 mL) = 87.9 g

Molality = # molsolute / # kg solvent

So I convert above answer to kg:
87.9 g * (1 kg/1000 g)=.0879 kg Benzene

M.W. Benzene: 122.12 g
This should be benzoic acid. That is the solute in the problem. Benzene is the solvent and you have already converted that to kg.
?mol Benzene :
7.58g benzene*(1 mol benzene/122.12g)=0.0621 mol Benzene.
Just convert the word benzene to benzoic acid. You used the molar mass of 122.12 correctly for benzoic acid. I suspect this is just a little confusion between the similarity of the two names.
Molality =
.0621 mol/.0879 kg =.706 mol/kg benzene
I would have written 0.0706 m (note the first 0 in front of the decimal AND the small m. Your answer is correct and there is noting wrong with writing it as you did but the other way avoids confusion that sometimes exists. Good work!
2. What is the molarity of a Ba(OH)2 solution if 15.48 mL of it are required to react with 25 mL of .303 M HCl solution? Show the balanced reaction in your calculations.

Sol:
2HCl(aq)+ Ba(OH)2(aq)-->BaCl2 + 2H2O(L)
Since you have used (aq) for everything on the left, I think you should have written (aq) after the BaCl2, also. You have it for everyting else so I suspect this is just an oversight.
(there are some exceptions when dealing with the formula: m1v1 = m2v2; I know to apply when you deal with the same compound. Here we have two. I'll use formula anyways as I see no way to proceed.
You COULD use a modified formula but that can get confusing. I recommend the following because it follows a logical path; i.e., calculate mols HCl, convert using the equation to mols Ba(OH)2, convert mols Ba(OH)2 to M by M = mols/L.
(2*.3030M HCl)*.025L = .0155M2
.0152M = m2 ??

When the coefficients for the balanced equation are not the same, as in this case, it isn't proper to use the formula m1v1 = m2v2. The best way to proceed is as follows:
mols HCl used in the titration is M x L = 0.303 x 0.025 L = 0.00758 mols HCl.
Now convert mols HCl to mols Ba(OH)2 using the balanced chemical equation. The numbers come from the coefficients in the equation.
0.00758 mols HCl x (1 mol Ba(OH)2/2 mols HCl) = 0.00379 mols Ba(0H)2.
Note that this is the same way you converted g to kg in the first problem; i.e., in this case mols HCl in the numerator of the first term cancel with mols HCl in the denominator of the second term and that leaves units of Ba(OH)2 for the answer.
Now convert mols Ba(OH)2 to M remembering that mols = M x L. Solving for M we have M = mols/L = 0.00379/0.01548 L = 0.245 M. Check my math. Check my work. I may not have rounded exactly. I hope this helps.

  1. 👍
  2. 👎
  3. 👁

Respond to this Question

First Name

Your Response

Similar Questions

  1. Math

    a 25.00 gram sample is placed in a graduated cylinder and then the cylinder is filled to the 50.0 mL mark with benzene. the mass of benzene and solid together is 58.80 gram. assuming that the solid is insoluble in benzene and that

  2. science

    the melting point of pure benzoic acid and pure naphthol are 122.5 and 123 degree Celsius respectively. you are given a pure sample that is known to be either pure benzoic acid or 2- naphthol. describe a procedure you will use to

  3. Ochem

    A student needed to determine the percent composition of a mixture that contained fluorene and benzoic acid. The student took 283 mg of the mixture and used the extraction method to separate the components. Then both compounds

  4. Chemistry

    The piperazine content of an impure commercial material can be determined by precipitating and weighing the diacetate: (C2H4)2N2H2 + 2 CH3COOH → (C2H4)2N2H2(CH3COOH)2 In one experiment, 0.3126 gram of the sample was dissolved in

  1. ap chemistry

    A 0.456 gram sample of an unknown mono- protic acid (let’s call it HZ) was dissolved in some water (you pick the amount). Then the acidic solution was titrated to the equivalence point with 32.5 mL of 0.174 M KOH. What is the

  2. Organic Chemistry

    An impure sample contains 0.95g of benzoic acid and 0.05 of salicylic acid. Solubilities in water of the two compounds are given in the following table. Solubility at 20 degrees (g/10mL) benzoic acid: 0.029 salicylic acid: 0.22

  3. Organic Chemistry - Extraction

    You have 100 mL of a solution of benzoic acid in water; the amount of benzoic acid in the solution is estimated to be about 0.30 g. The distribution coefficient of benzoic acid in diethyl ether and water is approximately 10.

  4. Chemistry

    The combustion reaction for benzoic acid C6H5CO2H(s) + 15 2 O2(g) → 7 CO2(g) + 3 H2O(ℓ) has ∆H0 = −3226.7 kJ/mole. Use Hess’s Law to calculate ∆H0 f for benzoic acid.

  1. Chemistry

    Benzene reacts with hot concentrated nitric acid dissolved in sulfuric acid to give chiefly nitrobenzene, C6H5NO2. A by-product is often obtained, which consists of 42.86% C, 2.40% H, and 16.67% N (by mass). The boiling point of a

  2. Chemistry

    Ascorbic acid (vitamin C) is a diprotic acid having the formula H2C6H6O6. A sample of a vitamin supplement was analyzed by titrating a 0.3252 g sample dissolved in water with 0.0284 M NaOH. A volume of 13.27 mL of the base was

  3. Chemistry Science

    A 15.0 g sample of an unknown compound is dissolved in 100.0 g of benzene. The boiling point is raised 2.67 oC above the boiling point of pure benzene. What is the molar mass of the sample? Kb for benzene is 2.67oC/m.

  4. Chemistry

    The combustion of benzoic acid releases 26.38kJ/g and is often used to calibrate a bomb calorimete. The combustion of 1.045g of benzoic acid caused the temperature of the calorimeter to rise by 5.985 degreeC.Using the same

You can view more similar questions or ask a new question.